Question

A sample of neon occupies a volume of 461 mL at STP. What will be the
volume of the neon when the pressure is reduced to 93.3 kPa?

Answer 1

Heya.......!!!!!

It's a question of Boyle's Law .

Given that :

=> P1 = 101.3 kPa
=> P2 = 93.3 kPa

=> V1 = 461 mL
=> V2 = ?

By equation of Boyle's Law

P1V1 = P2V2 .

=>> V2 = P1V1/P2
V2 = 101.3 x 461 /93.3
➡ V2 = 501 ml Neon .



Hope It Helps u ^_^

Answer 2

The absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system.

Mathematically, Boyle's law can be stated as:

    PV = k (Pressure multiplied by volume equals some constant k)

• Where  P is the pressure of the gas
• V is the volume of the gas
• k is a constant

For comparing the same substance under two different sets of conditions, the law can be usefully expressed as:

                      $P_{1} V_{1} = P_{2} V_{2}$

Given that ,

     $P_{1}$ = 101.3 kPa                    $V_{1}$ = 461 mL

     $P_{2}$ = 93.3 kPa                     $V_{2}$ = ?

Inserting these values in the equation of boyle's law we get ,

⇒     $P_{1} V_{1} = P_{2} V_{2}$

⇒            $V_{2}$ = $\frac{P_{1} V_{1} }{P_{2} }$

$V_{2}$ = $\frac{101.3 * 461}{93.3}$

➡ $V_{2}$ = 501 ml Neon .